Yes, that's precisely correct. You can't solve for a because a doesn't appear in any of the equations. In particular if b= c= d= 0, No matter what a is, the linear transformation is will take (a, 0, 0, 0) to (0, 0, 0, 0).
What does that tell you about one basis vector for the null space?

You also have b+ 4c= 0 so that c= -(1/4)b as well as 2b- 5d= 0 so that
d= (2/5)b. That seems to say that if you take b to be anything, c= -(1/4)b, d= (2/5)b and a to be anything, all the equations will be satisfied. What does that tell you about the dimension of the kernel? What is a basis for the kerne?

Quick question, I'm now trying to find the imagine space of this matrix, but I took the determinant of the matrix and it is equal to 0, doesn't that mean there is no image space? Because there is a row of 0's it can't span R^4, so there can't be a unique solution for all varaibles right?